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jupyter notebook showing calculation of the valuation
This commit is contained in:
narodnik
564
script/research/ec/valuation.ipynb
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564
script/research/ec/valuation.ipynb
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{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "6a512e0d",
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"metadata": {},
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"source": [
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"# Computing Valuation Manually\n",
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"\n",
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"We wish to compute $\\textrm{ord}_P(f)$ where $P = (2, 4)$ and $f = y - 2x$.\n",
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"\n",
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"$$E(\\mathbb{F}_11) : y^2 = x^3 + 4x$$\n",
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"\n",
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"Let $K[V] = K[x, y] / \\langle y^2 - x^3 - 4x \\rangle$ by the coordinate ring. $K(V)$ is the field of fractions for $K[V]$."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 5,
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"id": "5d554e0c",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"Fraction Field of Quotient of Multivariate Polynomial Ring in x, y over Finite Field of size 11 by the ideal (-x^3 + y^2 - 4*x)"
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]
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},
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"execution_count": 5,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"# Define our base polynomial ring over Z_11\n",
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"K.<x, y> = GF(11)[]\n",
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"# This is K(V)\n",
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"S = K.quotient(y^2 - x^3 - 4*x).fraction_field()\n",
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"S"
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]
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},
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{
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"cell_type": "markdown",
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"id": "69feb612",
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"metadata": {},
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"source": [
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"Because $P$ lies on the curve $E$, we can take the nontangent lines $x - 2, y - 4$ as a basis for the local curve. The intuition is that these lines describe the coordinate grid around $P$, and we can multiply them by any polynomial to get cosets of $K(V)$.\n",
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"\n",
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"More formally we can see this by noting that:\n",
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"\n",
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"$$(y - 4)(y + 4) = (x - 2)^3 - 5*(x - 2)^2 - 6(x - 2)$$\n",
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"\n",
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"So therefore any function on $E$ can be expressed in terms of $(x - 2)$ and $(y - 4)$."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 11,
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"id": "c430ae30",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"True"
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]
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},
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"execution_count": 11,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"X, Y = S(x), S(y)\n",
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"(Y - 4)*(Y + 4) == (X - 2)^3 - 5*(X - 2)^2 - 6*(X - 2)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "2515224a",
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"metadata": {},
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"source": [
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"Begin by expressing $f = y - 2x$ in terms of this basis.\n",
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"\\begin{align*}\n",
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"y &= y - 2x \\\\\n",
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" &= -2(x - 2) + 1(y - 4) \\\\\n",
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"\\mathbf{a} &= (-2, 1, 0) \\\\\n",
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"\\mathbf{b} &= (x - 2, y - 4, 1)\n",
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"\\end{align*}"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 10,
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"id": "bd415ef9",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"-2*x + y"
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]
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},
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"execution_count": 10,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"# Components for f\n",
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"a0, a1, a2 = -2, 1, 0\n",
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"# Our basis\n",
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"b0, b1, b2 = x - 2, y - 4, 1\n",
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"a0*b0 + a1*b1 + a2*b2"
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]
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},
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{
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"cell_type": "markdown",
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"id": "e3087005",
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"metadata": {},
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"source": [
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"Using the identity above for $(y - 4)(y + 4)$, we can see that\n",
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"\\begin{align*}\n",
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"(y - 4) &= \\frac{(x - 2)^3 - 5*(x - 2)^2 - 6(x - 2)}{(y + 4)} \\\\\n",
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" &= (x - 2) \\frac{(x - 2)^2 - 5*(x - 2) - 6}{(y + 4)}\n",
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"\\end{align*}\n",
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"So we can know that $b_1 = (E_f / E_g) b_0$, and can make this substitution in $f$.\n",
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"\\begin{align*}\n",
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"f &= -2 (x - 2) + 1 \\cdot \\frac{(x - 2)^3 - 5(x - 2)^2 - 6(x - 2)}{y + 4} \\\\\n",
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" &= \\frac{1}{(y + 4)}[(x - 2)^3 - 5(x - 2)^2 - 6(x - 2) - 2(x - 2)(y + 4)]\n",
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"\\end{align*}"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 13,
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"id": "9e2be279",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"True"
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]
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},
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"execution_count": 13,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"# and lets double check this\n",
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"B0 = X - 2\n",
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"Eg = Y + 4\n",
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"(B0^3 - 5*B0^2 - 6*B0 - 2*B0*Eg)/Eg == Y - 2*X"
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]
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},
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{
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"cell_type": "markdown",
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"id": "9dfe225e",
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"metadata": {},
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"source": [
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"Now lets quotient out $b_0 = (x - 2)$ from $f$ to get the first power of $k = 1$ for the uniformizer.\n",
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"\\begin{align*}\n",
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"f &= \\frac{(x + 2)^1}{(y + 4)}[(x - 2)^2 - 5(x - 2) - 6 - 2(y + 4)]\n",
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"\\end{align*}\n",
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"Notice now the constant term is $- 6 - 2(y + 4)$ which should be a multiple of $y - 4$ if we can continue extracting $b_0$ from the expression."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 14,
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"id": "249b0ed0",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"-2*y - 3"
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]
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},
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"execution_count": 14,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"-6 - 2*(y + 4)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 15,
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"id": "5894ffc4",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"-2*y - 3"
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]
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},
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"execution_count": 15,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"-2*(y - 4)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "dc5c6b41",
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"metadata": {},
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"source": [
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"Great so we continue. Instead of writing $-6 - 2(y + 4)$, lets instead write $-2(y - 4)$\n",
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"\\begin{align*}\n",
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"f &= \\frac{(x + 2)^1}{(y + 4)}[(x - 2)^2 - 5(x - 2) - 2(y - 4)] \\\\\n",
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" &= \\frac{(x + 2)^1}{(y + 4)}[(x - 2)^2 - 5(x - 2) - 2\\frac{(x - 2)^3 - 5(x - 2)^2 - 6(x - 2)}{y + 4}] \\\\\n",
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" &= \\frac{(x + 2)^1}{(y + 4)^2}[(x - 2)^2(y + 4) - 5(x - 2)(y + 4) - 2((x - 2)^3 - 5(x - 2)^2 - 6(x - 2))] \\\\\n",
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" &= \\frac{(x + 2)^2}{(y + 4)^2}[(x - 2)(y + 4) - 5(y + 4) - 2((x - 2)^2 - 5(x - 2) - 6)] \\\\\n",
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"\\end{align*}\n",
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"So $k = 2$, and lets evaluate the constant terms."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 16,
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"id": "7b343008",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"-5*y + 3"
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]
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},
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"execution_count": 16,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"-5*(y + 4) - 2*(-6)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 17,
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"id": "595834a3",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"-5*y - 2"
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]
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},
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"execution_count": 17,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"-5*(y - 4)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "8928eeda",
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"metadata": {},
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"source": [
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"Now there's a remainder left over so the expression terminates."
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]
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},
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{
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"cell_type": "markdown",
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"id": "5575757f",
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"metadata": {},
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"source": [
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"# Implementation Details\n",
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"\n",
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"Several points of interest:\n",
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"\n",
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"* We don't need to keep track of $g$ although it's done for completeness.\n",
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"* For the uniformizer part, we just need to keep track of $k$.\n",
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"* The inner expression can be optimized just by looking at the constant term when viewed from $(x - 2)$. Although we are not doing that."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 18,
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"id": "a0a83e30",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"(x - 2, y - 4, 1)"
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]
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},
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"execution_count": 18,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"b0, b1, b2"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 24,
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"id": "d042e184",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"(-2, 1, 0)"
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]
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},
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"execution_count": 24,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"# Decompose polynomial function into basis components\n",
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"def decomp(f):\n",
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" a0, r = f.quo_rem(b0)\n",
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" a1, r = r.quo_rem(b1)\n",
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" a2, r = r.quo_rem(b2)\n",
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" return a0, a1, a2\n",
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"\n",
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"f = y - 2*x\n",
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"decomp(f)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 23,
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"id": "c804233f",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"-x^3 + y^2 - 4*x"
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]
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},
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"execution_count": 23,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"Px, Py = (2, 4)\n",
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"EC_A = 4\n",
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"# Calculate substitution polynomials\n",
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"Ef = b0^2 + binomial(3,2)*Px*b0^1 + (3*Px^2 + EC_A)\n",
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"Eg = (y + Py)\n",
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"# Should be EC equation\n",
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"Eg*b1 - Ef*b0"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 27,
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"id": "cbbb63c7",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"(x^2 + 2*x - 2*y, y + 4)"
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]
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},
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"execution_count": 27,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"a0, a1, a2 = decomp(f)\n",
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"g = 1\n",
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"\n",
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"# Perform first reduction\n",
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"a0 = Eg*a0 + Ef*a1\n",
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"a1 = 0\n",
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"g *= Eg\n",
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"assert a1 == a2 == 0\n",
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"a0, g"
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]
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||||
},
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{
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||||
"cell_type": "markdown",
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"id": "9357c553",
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||||
"metadata": {},
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||||
"source": [
|
||||
"Set $k = 1$ since we have now factored out $(x - 2)$. Continue for second decomposition followed by reduction."
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||||
]
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},
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||||
{
|
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"cell_type": "code",
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"execution_count": 28,
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"id": "d5810612",
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"metadata": {},
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||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"(x + 4, -2, 0)"
|
||||
]
|
||||
},
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||||
"execution_count": 28,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
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||||
}
|
||||
],
|
||||
"source": [
|
||||
"a0, a1, a2 = decomp(a0)\n",
|
||||
"a0, a1, a2"
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||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "2a5b3561",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"as expected the remainder a2 is zero"
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||||
]
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||||
},
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||||
{
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||||
"cell_type": "code",
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||||
"execution_count": 31,
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||||
"id": "22671821",
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||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
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||||
"text/plain": [
|
||||
"True"
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||||
]
|
||||
},
|
||||
"execution_count": 31,
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||||
"metadata": {},
|
||||
"output_type": "execute_result"
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||||
}
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||||
],
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"source": [
|
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"U = (X - 2)\n",
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"F = (X + 4)*(X - 2) - 2*(Y - 4)\n",
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||||
"G = Y + 4\n",
|
||||
"U^1 * F / G == Y - 2*X"
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||||
]
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||||
},
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||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 32,
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||||
"id": "411c7216",
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"(-2*x^2 + x*y + 4*y, y^2 - 3*y + 5)"
|
||||
]
|
||||
},
|
||||
"execution_count": 32,
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||||
"metadata": {},
|
||||
"output_type": "execute_result"
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||||
}
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||||
],
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"source": [
|
||||
"# Second reduction\n",
|
||||
"a0 = Eg*a0 + Ef*a1\n",
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||||
"a1 = 0\n",
|
||||
"g *= Eg\n",
|
||||
"assert a1 == a2 == 0\n",
|
||||
"a0, g"
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||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "de5d1ed9",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"Set $k = 2$. Now decompose and check remainder is zero again before performing reduction."
|
||||
]
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||||
},
|
||||
{
|
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"cell_type": "code",
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"execution_count": 33,
|
||||
"id": "eceb52f7",
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"(-2*x + y - 4, -5, 5)"
|
||||
]
|
||||
},
|
||||
"execution_count": 33,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"a0, a1, a2 = decomp(a0)\n",
|
||||
"a0, a1, a2"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"id": "daab41c5",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"Now the remainder is 5 so the algorithm stops. Our final valuation is $k = 2$."
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 44,
|
||||
"id": "1c654c46",
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": [
|
||||
"# Convert our function to the function field K(V)\n",
|
||||
"original_f = Y - 2*X\n",
|
||||
"k = 2\n",
|
||||
"\n",
|
||||
"f = a0*b0 + a1*b1 + a2*b2\n",
|
||||
"fprime = b0^k * f/g\n",
|
||||
"assert fprime == S(original_f)\n",
|
||||
"assert g(Px, Py) != 0\n",
|
||||
"assert f(Px, Py) != 0\n",
|
||||
"assert b0(Px, Py) == 0\n"
|
||||
]
|
||||
}
|
||||
],
|
||||
"metadata": {
|
||||
"kernelspec": {
|
||||
"display_name": "SageMath 9.6",
|
||||
"language": "sage",
|
||||
"name": "sagemath"
|
||||
},
|
||||
"language_info": {
|
||||
"codemirror_mode": {
|
||||
"name": "ipython",
|
||||
"version": 3
|
||||
},
|
||||
"file_extension": ".py",
|
||||
"mimetype": "text/x-python",
|
||||
"name": "python",
|
||||
"nbconvert_exporter": "python",
|
||||
"pygments_lexer": "ipython3",
|
||||
"version": "3.10.5"
|
||||
}
|
||||
},
|
||||
"nbformat": 4,
|
||||
"nbformat_minor": 5
|
||||
}
|
||||
Reference in New Issue
Block a user